Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Sujay Group Title

You are a farmer, and you are trying to make the most efficient use of your resources. You have invested enough capital for a barn house which is 40 feet by 70 feet (its base) and 100 feet of fencing for your animals. You decide to have an outdoor lot for your animals. Your goal is to have the lot with the most area with the least amount of fencing (optimize). The lot must be rectangular, and one side of the rectangular may be a wall from the barn. A picture will be posted below.

  • one year ago
  • one year ago

  • This Question is Closed
  1. Sujay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1356887557708:dw|

    • one year ago
  2. AnimalAin Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The area of your pen is the length times the width. Since you are going to have three sides of the pen enclosed by your fencing (the last side will be from the barn), we have a second specification that must be met, the length plus twice the width is 100. Rearrange the second equation to solve for a single variable (either l or w) and substitute into the first equation. You will get a quadratic function; find the maximum value, and you have the width (or maybe length) that corresponds to the maximum area. One more substitution gives the dimensions of the pen..

    • one year ago
  3. AnimalAin Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[A=lw\]\[100=l+2w \implies l=100-2w \implies\]\[A=lw=A(w)=(100-2w)w=-2w^2+100\]To find the maximum (or minimum in other cases) of a quadratic function f(x)=ax^2+bx+c, we take the value of the function at x=-b/(2a). In this case, that gives us\[w=\frac{-100}{2(-2)}=25 \implies l=50\]by substitution. So our enclosure is 25' by 50', against the side of the barn that is 70' long.

    • one year ago
  4. Sujay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Wow thanks, I totally forgot we don't need calculus for this problem.

    • one year ago
  5. Sujay Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Can anyone do it using optimization?

    • one year ago
  6. AnimalAin Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The optimization problem is the same, except when you get A(w), you take the derivative and set it to zero. Since the second derivative is negative at that value of w, the function has a maximum at that point.\[A(w)=−2w^2+100w \implies A'(w)=-4w+100=0 \implies w=25\]\[A"(w)=-4<0 \implies maximum~at~critical~value\]

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.